Let $C$ be an axis-parallel orthogonal polygon with a finite number of sides. Define an *anti-rectangle* in $C$ as a set of small squares in $C$, such that no two of them are covered by a single large rectangle in $C$. Define a *maximum anti-rectangle* as an anti-rectangle that contains a maximum number of squares. For example, the following L-shape has a maximum anti-rectangle with 2 squares:

![L-shape][1]

The following C-shape has a maximum anti-rectangle with 3 squares:

![C-shape][2]

And the following shape has a maximum anti-rectangle with 5 squares:

![C-C-shape][3]

When I try to build an anti-rectangle, I usually start with putting squares in the corners of $C$, because intuitively, the corners have the least chance of being covered by a rectangle. This raises the following conjecture:

**For every $C$, there is a maximum anti-rectangle, which contains a corner square.**

(- a *corner square* is a square with at least two adjacent sides that are in the boundary of $C$).

I found in [Chaikan et al (1981)](http://dx.doi.org/10.1137/0602042) a proof that the conjecture is true when $C$ is *linearly-convex* (- contains every vertical or horizontal line that connects two of its points; like the L-shape above). 

Their proof is not valid when $C$ is not linearly-convex, but I also cannot find a counter-example. What do you say?

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CONCLUSION: Many thanks to all repliers. The conjecture is false when the polygon may have holes (as shown by [dmotorp][4]). It is true when the polygon is hole-free (as proven by [Nick Gill][5] and [mhum][6]).

UPDATE: I cited this thread in the following [working paper][7]. I hope it will be accepted.

  [1]: https://i.sstatic.net/FjvTp.png
  [2]: https://i.sstatic.net/17bKQ.png
  [3]: https://i.sstatic.net/vYUhL.png
  [4]: https://mathoverflow.net/a/148563/34461
  [5]: https://mathoverflow.net/a/148815/34461
  [6]: https://mathoverflow.net/a/149020/34461
  [7]: http://econpapers.repec.org/paper/biuwpaper/2014-01.htm